Question

Test the following series for convergence or divergence. Decide for yourself which test is easiest to use, but don’t forget the preliminary test. Use the facts stated above when they apply

$\mathbf{\sum }_{\mathbf{n}\mathbf{=}\mathbf{1}}^{\mathbf{\infty }}\frac{\mathbf{n}\mathbf{-}\mathbf{1}}{\mathbf{(}\mathbf{n}\mathbf{+}\mathbf{2}\mathbf{)}\mathbf{(}\mathbf{n}\mathbf{+}\mathbf{3}\mathbf{)}}$

Short answer

The series is divergence.

Step-by-step solution

Significance of alternating series

If the terms' absolute values gradually decline to zero, or if $\mathbf{\left|}{\mathit{a}}_{\mathbf{n}\mathbf{+}\mathbf{1}}\mathbf{\right|}\mathbf{\le }\mathbf{\left|}{\mathit{a}}_{\mathbf{n}}\mathbf{\right|}$and$\underset{\mathbf{n}\mathbf{\to }\mathbf{\infty }}{\mathbf{l}\mathbf{i}\mathbf{m}}{\mathit{a}}_{\mathbf{n}}\mathbf{=}\mathbf{0}$, then an alternating series converges.

Given series

Consider the following series

$\sum _{\mathrm{n}=1}^{\infty }\frac{\mathrm{n}-1}{(\mathrm{n}+2)(\mathrm{n}+3)}$

Use one of the well-known tests to see if the series is convergent.

Apply comparison test

The special comparison test is the most appropriate test for this series, so let's begin by using it on the following series:

To see that for$n\ge 2,\frac{n-1}{(n+2)(n+3)}>0$

The specific comparison test for can now be performed as follows:

$$\begin{gathered} \underset{\mathrm{n}\to \infty }{\lim }\frac{a_n}{b_n}=\underset{\mathrm{n}\to \infty }{\lim }\frac{{\displaystyle \frac{n-1}{(n+2)(n+3)}}}{{\displaystyle \frac{1}{n}}} \\ \underset{\mathrm{n}\to \infty }{\lim }\frac{{\mathrm{a}}_{\mathrm{n}}}{{\mathrm{b}}_{\mathrm{n}}}=\underset{\mathrm{n}\to \infty }{\lim }\frac{n\left(n-1\right)}{(n+2)(n+3)} \end{gathered}$$

Simplify by dividingn2

Because this is an indeterminate form of limit, we shall divide both the denominator and the numerator by $n^2$resulting in:

$$\begin{gathered} \underset{\mathrm{n}\to \infty }{\lim }\frac{a_n}{b_n}=\underset{\mathrm{n}\to \infty }{\lim }\frac{\left(1-{\displaystyle \frac{1}{n}}\right)}{\left(1+{\displaystyle \frac{2}{n}}\right)\left(1+{\displaystyle \frac{3}{n}}\right)} \\ =\frac{\left(1+{\displaystyle \frac{1}{\infty }}\right)}{\left(1+{\displaystyle \frac{2}{\infty }}\right)\left(1+\frac{3}{\infty }\right)} \\ =1 \end{gathered}$$

According to the special comparison test,$a_n$ is diverges